Givet en sorteret række af distinkte positive heltal udskriv alle tripletter, der danner geometrisk progression med integral fælles forhold.
En geometrisk progression er en række tal, hvor hvert led efter det første findes ved at gange det foregående med et fast tal, der ikke er nul, kaldet det fælles forhold. For eksempel er sekvensen 2 6 18 54... en geometrisk progression med fælles forhold 3.
Eksempler:
Input: arr = [1 2 6 10 18 54] Output: 2 6 18 6 18 54 Input: arr = [2 8 10 15 16 30 32 64] Output: 2 8 32 8 16 32 16 32 64 Input: arr = [ 1 2 6 18 36 54] Output: 2 6 18 1 6 36 6 18 54
Ideen er at starte fra det andet element og fastgøre hvert element som midterelement og søge efter de to andre elementer i en triplet (et mindre og et større). For at et element arr[j] skal være midt i geometrisk progression skal der eksistere elementer arr[i] og arr[k], således at -
arr[j] / arr[i] = r and arr[k] / arr[j] = r where r is an positive integer and 0 <= i < j and j < k <= n - 1
Nedenfor er implementeringen af ovenstående idé
strengfunktioner i javaC++
// C++ program to find if there exist three elements in // Geometric Progression or not #include
// Java program to find if there exist three elements in // Geometric Progression or not import java.util.*; class GFG { // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. static void findGeometricTriplets(int arr[] int n) { // One by fix every element as middle element for (int j = 1; j < n - 1; j++) { // Initialize i and k for the current j int i = j - 1 k = j + 1; // Find all i and k such that (i j k) // forms a triplet of GP while (i >= 0 && k <= n - 1) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while (i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0 && arr[j] / arr[i] == arr[k] / arr[j]) { // print the triplet System.out.println(arr[i] +' ' + arr[j] + ' ' + arr[k]); // Since the array is sorted and elements // are distinct. k++ ; i--; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if(i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0) { if(i >= 0 && arr[j] / arr[i] < arr[k] / arr[j]) i--; else k++; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if (i >= 0 && arr[j] % arr[i] == 0) k++; else i--; } } } // Driver code public static void main(String[] args) { // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; int arr[] = {1 2 4 16}; // int arr[] = {1 2 3 6 18 22}; int n = arr.length; findGeometricTriplets(arr n); } } // This code is contributed by Rajput-Ji
# Python 3 program to find if # there exist three elements in # Geometric Progression or not # The function prints three elements # in GP if exists. # Assumption: arr[0..n-1] is sorted. def findGeometricTriplets(arr n): # One by fix every element # as middle element for j in range(1 n - 1): # Initialize i and k for # the current j i = j - 1 k = j + 1 # Find all i and k such that # (i j k) forms a triplet of GP while (i >= 0 and k <= n - 1): # if arr[j]/arr[i] = r and # arr[k]/arr[j] = r and r # is an integer (i j k) forms # Geometric Progression while (arr[j] % arr[i] == 0 and arr[k] % arr[j] == 0 and arr[j] // arr[i] == arr[k] // arr[j]): # print the triplet print( arr[i] ' ' arr[j] ' ' arr[k]) # Since the array is sorted and # elements are distinct. k += 1 i -= 1 # if arr[j] is multiple of arr[i] # and arr[k] is multiple of arr[j] # then arr[j] / arr[i] != arr[k] / arr[j]. # We compare their values to # move to next k or previous i. if(arr[j] % arr[i] == 0 and arr[k] % arr[j] == 0): if(arr[j] // arr[i] < arr[k] // arr[j]): i -= 1 else: k += 1 # else if arr[j] is multiple of # arr[i] then try next k. Else # try previous i. elif (arr[j] % arr[i] == 0): k += 1 else: i -= 1 # Driver code if __name__ =='__main__': arr = [1 2 4 16] n = len(arr) findGeometricTriplets(arr n) # This code is contributed # by ChitraNayal
// C# program to find if there exist three elements // in Geometric Progression or not using System; class GFG { // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. static void findGeometricTriplets(int []arr int n) { // One by fix every element as middle element for (int j = 1; j < n - 1; j++) { // Initialize i and k for the current j int i = j - 1 k = j + 1; // Find all i and k such that (i j k) // forms a triplet of GP while (i >= 0 && k <= n - 1) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while (i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0 && arr[j] / arr[i] == arr[k] / arr[j]) { // print the triplet Console.WriteLine(arr[i] +' ' + arr[j] + ' ' + arr[k]); // Since the array is sorted and elements // are distinct. k++ ; i--; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if(i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0) { if(i >= 0 && arr[j] / arr[i] < arr[k] / arr[j]) i--; else k++; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if (i >= 0 && arr[j] % arr[i] == 0) k++; else i--; } } } // Driver code static public void Main () { // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; int []arr = {1 2 4 16}; // int arr[] = {1 2 3 6 18 22}; int n = arr.Length; findGeometricTriplets(arr n); } } // This code is contributed by ajit.
<script> // Javascript program to find if there exist three elements in // Geometric Progression or not // The function prints three elements in GP if exists // Assumption: arr[0..n-1] is sorted. function findGeometricTriplets(arrn) { // One by fix every element as middle element for (let j = 1; j < n - 1; j++) { // Initialize i and k for the current j let i = j - 1 k = j + 1; // Find all i and k such that (i j k) // forms a triplet of GP while (i >= 0 && k <= n - 1) { // if arr[j]/arr[i] = r and arr[k]/arr[j] = r // and r is an integer (i j k) forms Geometric // Progression while (i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0 && arr[j] / arr[i] == arr[k] / arr[j]) { // print the triplet document.write(arr[i] +' ' + arr[j] + ' ' + arr[k]+'
'); // Since the array is sorted and elements // are distinct. k++ ; i--; } // if arr[j] is multiple of arr[i] and arr[k] is // multiple of arr[j] then arr[j] / arr[i] != // arr[k] / arr[j]. We compare their values to // move to next k or previous i. if(i >= 0 && arr[j] % arr[i] == 0 && arr[k] % arr[j] == 0) { if(i >= 0 && arr[j] / arr[i] < arr[k] / arr[j]) i--; else k++; } // else if arr[j] is multiple of arr[i] then // try next k. Else try previous i. else if (i >= 0 && arr[j] % arr[i] == 0) k++; else i--; } } } // Driver code // int arr[] = {1 2 6 10 18 54}; // int arr[] = {2 8 10 15 16 30 32 64}; // int arr[] = {1 2 6 18 36 54}; let arr = [1 2 4 16]; // int arr[] = {1 2 3 6 18 22}; let n = arr.length; findGeometricTriplets(arr n); // This code is contributed by avanitrachhadiya2155 </script>
Produktion
1 2 4 1 4 16
Tidskompleksitet af ovenstående opløsning er O(n2) som for hvert j, vi finder i og k i lineær tid.
Hjælpeplads: O(1) da vi ikke brugte ekstra plads.